All Questions
Tagged with calculus acceleration
96
questions
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Differential Equation & MacLaurin Series for Newton’s Second Law
I am currently working with a differential equation, where I think I need to take the derivative of $ma$ (corrected as per comment). I am trying to write $F = ma$ as a MacLaurin series and eventually ...
- 151
0
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1
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Acceleration function of position and time
I have an acceleration function in python with position and time parameters and returns the acceleration value.
I need the end velocity at a position ,start velocity is zero.
how to calculate this ...
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3
answers
230
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Are acceleration and velocity simultaneous? [closed]
I would think yes because, if a rope tied to a swinging rock breaks, the rock flies off in the direction that is perpendicular to the direction of the last instant of the acceleration. The ...
- 71
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2
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85
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Kinematic displacement: why not represent higher order rates of change?
I understand that the equation for kinematic displacement is:
$x = v_{0x}t+\frac{1}{2}a_xt^2$
Perhaps my understanding is naive, but it seems like this leaves out higher order rates of change. Why ...
- 111
1
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2
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2k
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Acceleration as a function of position and time
I know if you have an acceleration as a function of $t$, $a(t)$, to find the velocity you simply integrate $a(t)$ with respect to $t$. Moreover, if the acceleration was a function of position, $a(x)$, ...
- 33
1
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2
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In the equation: $a = dv/dt$ , is $dt$ the time taken to achieve that instantaneous acceleration?
If you solve for $dt$ from $a = \frac{dv}{dt}$ , is it the time taken to to achieved that instantaneous acceleration?
$a$ : acceleration
$v$ : velocity
$t$ : time
- 87
9
votes
4
answers
2k
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Can I find the acceleration or velocity when my displacement-time graph is discontinuous?
Today, I encountered the problem where I was asked to find the velocity and acceleration from displacement-time graph but the displacement-time graph was discontinuous. So I am unable to find the ...
- 123
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Trying to prove that the expression for the radial component of the acceleration is equal to $\mathbf v\cdot \mathbf v/r$
I am trying to prove that the normal component of acceleration of a particle undergoing a curvilinear motion is equal to
$\mathbf v\cdot \mathbf v/r$.
Here $\mathbf v$ is the velocity of the particle ...
- 529
12
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4
answers
3k
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Integrating acceleration - wrong choice of bounds in textbooks?
I've noticed in my physics textbook (and in a lot of other popular sources), that the process of integrating non-constant acceleration to get to a velocity formula, the integrating bounds imposed on ...
- 221
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4
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When the direction of a movement changes, is the object at rest at some time?
The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).
Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, ...
user229097
5
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2
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How does instantaneous velocity or acceleration have any other numerical value than 0? [duplicate]
Instantaneous velocity is defined as the limit of average velocity as the time interval ∆t becomes infinitesimally small. Average velocity is defined as the change in position divided by the time ...
- 153
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6
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Question about derivation of kinematics equations
Apologies if this has been asked before, but I browsed the sub and couldn't find something specific.
I understand the derivation for one of the equations as follows:
\begin{gather}
\frac{dv}{dt} = a ...
- 69
1
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5
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152
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Equation of distance and time
How is this equation derived?
$$r = r_0 + ut + at²/2$$
where $r_0$ is the initial position of particle and $r$ is the position of the particle after all the motion it has undergone, $a$ and $t$ ...
- 131
2
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5
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Acceleration and motion can be in different direction?
I'm not getting what acceleration concept is and how it relates to motion and how motion and acceleration can be in different direction? And what's behind the concept of negative and positive ...
- 31
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2
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Intuition of Distance covered when accelerating [duplicate]
When you're moving at $5$ m/s for $1$ second, you have traveled $5$ m.
When you're moving at $5$ m/s (initial velocity) and you accelerate $2$ m/s for $1$ second, you have traveled $5$ m + $1$ m (...
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